1
Dipanjan Das
·2010-05-28 18:42:36
1)
Write:
\int e^x\left(lnx+\frac{1}{x^2}\right)dx=\int e^x\left(lnx+\frac{1}{x}\right)dx+\int e^x\left(-\frac{1}{x}+\frac{1}{x^2}\right)dx
1
Manmay kumar Mohanty
·2010-05-28 21:01:42
Q2. u hav to take sinx = 2tan x1 + tan2x not as u mentioned in HINT
1
Manmay kumar Mohanty
·2010-05-28 22:03:17
Q3)
\int \frac{(x-1)}{(x+1)\sqrt{x^{3}+x^{2}+x}}dx
\int \frac{(x^{2}-1)}{(x+1)^{2}\sqrt{x^{3}+x^{2}+x}}dx
\int \frac{(x^{2}-1)}{(x^{2}+2x+1)\sqrt{x^{3}+x^{2}+x}}dx
\int \frac{(x^{2}-1)}{x.x(x+2+\frac{1}{x})\sqrt{x+1+\frac{1}{x}}}dx
\int \frac{(1-\frac{1}{x^{2}})}{(x+2+\frac{1}{x})\sqrt{x+1+\frac{1}{x}}}dx [ the one that was x2 in deno. divide it in the nume. ]
Put x + 1 + 1x = t
→ 1 - 1x2dx = dt
\Rightarrow \int \frac{dt}{(t+1)\sqrt{t}}
put t = u2 → dt = 2u du
\Rightarrow \int \frac{2udu}{(u^{2}+1)u}\Rightarrow 2\int \frac{du}{1+u^{2}}
= 2tan^{-1}u+c=2tan^{-1}\sqrt{x+1+\frac{1}{x}}+c
11
Devil
·2010-05-28 23:27:07
2) Write the integral as
\int \frac{(sin^2\frac{x}{2}+cos^2\frac{x}{2})dx}{5(sin^2\frac{x}{2}+cos^2\frac{x}{2})+8sin\frac{x}{2}cos\frac{x}{2}}
Dvide numerator and denom by cos^2\frac{x}{2} then take tan\frac{x}{2}=k.
That's enough - I suppose.
39
Pritish Chakraborty
·2010-05-29 03:29:43
Abey yaar... you posted this in my cb aur mujhe kuch karne ka mauka tak nahi diya!! LOL